My Situation
- I have a N rectangles
- The rectangles all have the same shape (for example 2 inches wide x 1 inch tall) – Let’s refer to this size as Sw and Sh for the width and height
- I want to position these rectangles in a grid such that the rects completely on top and next to each other – like what you would see in a spreadsheet
- What I need is this: Given N, Sw, and Sh what are the number of rows (R) and columns (C) that would stack these rects into the most square-like arrangement possible
- It is understood that R & C may provide more cells than in needed (for example if N=15,Sw=1,Sh=1 then R=4,C=4 yielding 16 ‘slots’ for 15 rectangles – that is OK.
- If Sw=Sh then my humble math skills are enough – when they rectangles have differing widths and heights – well frankly that’s beyond me.
Some Notes
- Yes I have read this question: Stacking rectangles to take as little space as possible and no it did not help. Also it isnt the same question. That question is about rectangles that could be of different sizes, in this question the rectangles have the same size
- Yes I have searched on wolfram.com, etc and no luck there
- I don’t have a strong math background so I the way I phrasing this problem may itself be preventing me from finding the answer – I’ve tried related searches relating to tiling, dissecting, decomposing, and not had any success there either
Some examples
the * indicates the edges of the rects the | indicates that a cell is 'filled-in' Notice that not all R*C cells are filled in, but only and exactly N cells IF N=1, Sw=2, Sh=1 THEN R=1, C=1 ******** *||||||* ******** IF N=2, Sw=2, Sh=1 THEN R=2, C=1 ******** *||||||* ******** *||||||* ******** IF N=3, Sw=2, Sh=1 THEN R=2, C=2 *************** *||||||* * *************** *||||||*||||||* *************** IF N=4, Sw=2, Sh=1 THEN R=2, C=2 *************** *||||||*||||||* *************** *||||||*||||||* *************** IF N=5, Sw=2, Sh=1 THEN R=3, C=2 *************** *||||||* * *************** *||||||*||||||* *************** *||||||*||||||* *************** Implementation of AaronofTomorrow’s answer
# Implementation of AaronofTomorrow's answer # implemented in python 2.6 # reasonable output # works in constant time import math def f( N, Sw, Sh ) : cols = math.sqrt( float(N) * float(Sh) / float(Sw) ) cols = round(cols) rows = float(N) / float(cols) rows = math.ceil(rows) return (int(cols),int(rows)) Another implementation inspired by Will’s answer (Updated on 2008-12-08) – this is the one I finally used
# Another implementation inspired by Will's answer # implemented in python 2.6 # reasonable output - a bit better in yielding more squarelike grids # works in time proportional to number of rects # # strategy used it to try incrementaly adding a rect. # if the resulting rect requires more space then two # possibilities are checked - adding a new row or adding a new col # the one with the best aspect ratio (1:1) will be chosen def g( N, Sw, Sh ) : slope = float(Sh)/float(Sw) cols = 1 rows = 1 for i in xrange( N ) : num_to_fit =i+1 allocated_cells= cols* rows if ( num_to_fit <= allocated_cells ) : pass # do nothing else : hc,wc = float(Sh * rows), float(Sw * (cols+1)) hr,wr = float(Sh * (rows+1)), float(Sw * cols) thetac = math.atan( hc/wc) thetar = math.atan( hr/wr) alpha = math.pi/4.0 difr = abs(alpha-thetar) difc = abs(alpha-thetac) if ( difr < difc ) : rows = rows +1 else: cols = cols + 1 return (cols,rows)
Building on Will Dean’s response, find the derivative of his formula (with respect to nCols):
-N*Sh / nCols + Sw
Then set it to 0 and solve for nCols, which gives:
nCols = sqrt(N * Sh / Sw)
Round that and you should have the optimum number of columns:
cols = round(sqrt(N * Sh / Sw))
rows = ceil(N / cols)